Question
Read the passage given below and answer the following questions from 1 to 5.
When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below:
v = u + at
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
 2a s = v2 – u2
Where u is the initial velocity of the object which moves with uniform acceleration a for Time t, v is the final velocity and s is the distance travelled by the object in time.
  1. equation of motions are applicable to motion with
  1. uniform acceleration
  2. non uniform acceleration
  3. constant velocity
  4. none of these
  1. There are 4 equation of motion. True or false?
  1. True
  2. False
  1. The brakes applied to a car produce an acceleration of 10 m/s2 in the opposite direction to the motion. If the car takes 1 s to stop after the application of brakes, calculate the distance traveled during this time by car.
  1. An object is dropped from a tower falls with a constant acceleration of 10 m/s2. Find its speed 10 s after it was dropped.
  1. A bullet hits a Sand box with a velocity of 10 m/s and penetrates it up to a distance of 5 cm. Find the deceleration of the bullet in the sand box

Answer

  1. (a) uniform acceleration
  2. (b) False
  3. Here in this problem,

v = 0

a = -10 m/s2 (as acceleration is retarding)

t = 1 sec.

To find: distance travelled

Solution: using kinematic equation

v = u + at

0=  u + -10 × 1

u = 10 m/s

Therefore distance is given by

$\text{S}=\text{ut}+\frac{1}{2}\text{at}^2$

$\text{S}=10\times1-\Big(\frac{1}{2}\Big)\times10\times1^2$

s = 5m

  1. Here in this problem,

u = 0

a = 10 m/s2(as acceleration is in the direction of gravity)

t = 10 sec.

To find: final velocity after 10 second

Solution: using kinematic equation

v = u + at

v = 0 + 10 × 10

v = 100 m/s

  1. Here in this problem,

v = 0(as bullet is going to stop)

u = 10 m/s

s = 5m.

To find: deceleration of the bullet

Solution: using kinematic equation

2a s = v2 – u2

2 × a × 5 = 0- 102

10a = -100

$\text{A}=\frac{100}{10}$

a = -10m/s2. Negative sign indicates that it is deceleration.

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